Proof of Natural Number Inequality: a < b < c

[cragar]

Homework Statement

prove that if a; b; c are natural
numbers and if a < b and b < c, then a < c.

Some axioms we are allowed to use is if a<b then there exists a natural number e
such that a+e=b.

The Attempt at a Solution

If a<b then there is a natural number x such that a+x=b,
if b<c then there exists a natural number y such that b+y=c,
Now since b= a+x then a+x+y=c and since x+y is a another natural number, call it z then
a+z=c which implies a<c.
Is this the correct way to go about it.

 

[maajdl]

Right.